Re: state and file perms confusion

From perldoc:

Because the mode contains both the file type and its permissions, you 
+should mask off the file type portion and (s)printf using a "%o" if y
+ou want to see the real permissions.

    $mode = (stat($filename))[2];
    printf "Permissions are %04o\n", $mode & 07777;
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You've masked off the file type with $mode = $mode & 07777;, so you'll get 0644 (octal) from the script above when your $mode equals 420 (decimal), since the printf above makes the conversion from decimal to octal.

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Re^2: state and file perms confusion by neilwatson (Priest) on Aug 04, 2012 at 01:51 UTC
I'm missing something fundamental. `print "mode=".$mode."\n"; printf "printf are %04o\n", $mode; printf "Permissions are %04o\n", $mode & 07777; ..produces mode=33188 printf are 100644 Permissions are 0644` [download] What am I seeing? How can store the number in a scalar rather than printing it? Neil Watson watson-wilson.ca	[reply] [d/l]
Re^3: state and file perms confusion by Kenosis (Priest) on Aug 04, 2012 at 02:18 UTC
What am I seeing? `33188 decimal = 100644 octal (first printf) 07777 octal = 0000111111111111 binary (mask)-\| +- & (bitwise AND) 33188 decimal = 1000000110100100 binary (mode)-\| 0644 octal = 0000000110100100 binary (printf after mask) ^^^^ \| + - file type masked out` [download] How can store the number in a scalar rather than printing it? `use Modern::Perl; my $mode = 33188; my $modeResult = sprintf '%04o', $mode & 07777; say $modeResult; # prints 0644` [download]	[reply] [d/l] [select]
Re^3: state and file perms confusion by Tanktalus (Canon) on Aug 04, 2012 at 13:55 UTC
There aren't a lot of use cases for storing an octal "number" in a scalar. There are some, but they're all to do with formatting output, not actual use as a number. If your goal is to use this for chmod, you have to realise that it won't work. If you do this: `$mode = sprintf "%o", 420; # $mode = 644 chmod $mode, $file;` [download] you'll actually be doing the same thing as this: `chmod 01204, $file; # 01204 octal == 644 decimal` [download] Definitely not what you wanted. Leave the number as-is if you plan on using it in chmod. Note that the following lines are equivalent: `# 1. octal number chmod 0644, $file; # 2. decimal number chmod 420, $file; # 3. number in a scalar $mode = 0644; chmod $mode, $file; # 4. number in a scalar (from decimal) $mode = 420; chmod $mode, $file; # 5. number with bitmasks $mode = 0666 & ~022; # like umask chmod $mode, $file; # 6. decimal with bitmasks $mode = 438 & ~022; # 438 == 0666 chmod $mode, $file;` [download] So, to answer your question, the number is already in the scalar. You don't actually need to do anything about it. And, if you use `$mode = sprintf "%o", $mode;`, remember what you just did: you took a number, you printf'd it into a string, and put that back in $mode. So $mode no longer carries a number but a string. Perl will try to force it back to a number if that's what you try to do, but it won't be the number you started with.	[reply] [d/l] [select]